The length of a rubber cord is l₁ meters when the tension is  4N and l₂ meters when the the tension is 5N. The length in meters when the tension is 9 N is 

According to the statement in the question, a rubber cord's length changes to l1 when the tension is 4 N and to l2 when the tension is 5 N. When the tension reaches 9N, we are asked to determine the length of the same cord.

The Young's modulus for a wire's longitudinal extension is provided by the formula $\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A}∆\mathrm{L}}$

For the tension of 4N,

$\mathrm{Y}=\frac{4\times \mathrm{L}}{\mathrm{A}\mathit{(}{l}_1-\mathrm{L})}\mathrm{\_\_\_\_\_\_\_\_\_\_}\left(1\right)$

for the tension 5N,

$\mathrm{Y}=\frac{5\times \mathrm{L}}{\mathrm{A}(l_2-\mathrm{L})}\mathrm{\_\_\_\_\_\_\_\_\_\_}\left(2\right)$

for the tension 9N,

$\mathrm{Y}=\frac{9\times \mathrm{L}}{\mathrm{A}(l_3-\mathrm{L})}\mathrm{\_\_\_\_\_\_\_\_\_\_}\left(3\right)$

combining equations 1 and 2

$\frac{9\times \mathrm{L}}{\mathrm{A}(l_3-\mathrm{L})}=\frac{4\times \mathrm{L}}{\mathrm{A}(l_1-\mathrm{L})}$

$9(l_1-\mathrm{L})=4(l_3-\mathrm{L})$

$4l_3+5\mathrm{L}=9l_1$

multiplying both sides by 4,

$16l_3+20\mathrm{L}=36l_1\mathrm{\_\_\_\_\_\_\_\_\_}\left(4\right)$

similarly combining equations 2 and 3

$\frac{9\times \mathrm{L}}{\mathrm{A}(l_3-\mathrm{L})}=\frac{5\times \mathrm{L}}{\mathrm{A}(l_2-\mathrm{L})}$

$9(l_2-\mathrm{L})=5(l_3-\mathrm{L})$

$5l_3+4L=9l_2$

multiplying both sides by 5

$25l_3+20L=45l_2\_\_\_\_\_\_\_\_\_\left(5\right)$

subtracting equation 4 from 5,

$9l_3=45l_2-36l_1$

$l_3=\frac{45}{9}{l}_2-\frac{36}{9}{l}_1$

$l_3=5l_2-4l_1$

Hence the correct answer is $5l_2-4l_1.$